Cambridge A Level Mathematics 9709 — 2017 May/June Paper 4 · Variant 2

2 A 5 m O B D 30Å 6 m C The diagram shows a wire ABCD consisting of a straight part AB of length 5 m and a part BCD in the shape of a semicircle of radius 6 m and centre O. The diameter BD of the semicircle is horizontal and AB is vertical. A small ring is threaded onto the wire and slides along the wire. The ring starts from rest at A. The part AB of the wire is rough, and the ring accelerates at a constant rate of 2.5 m s−2 between A and B. (i) Show that the speed of the ring as it reaches B is 5 m s−1. [1] ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ ........................................................................................................................................................ The part BCD of the wire is smooth. The mass of the ring is 0.2 kg. (ii) (a) Find the speed of the ring at C, where angle BOC = 30Å. [4] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ (b) Find the greatest speed of the ring. [2] ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................ ................................................................................................................................................

Mark scheme: 2(i) 2 2.5 5 = × × v (ms–1) B1 AG Using 2 2 2 = + v u as Total: 1 2(ii)(a) M1 Attempting PE loss or KE gain PE loss = 0.2 × 10 × 6 sin 30 [= 6] and KE gain = 0.5 × 0.2 × (v 2 – 5 2) A1 Both PE and KE correct both unsimplified [6 = 0.1(v 2 – 52)] M1 PE loss = KE gain (3 terms) v 2 = 85 → v = 9.22 ms–1 A1 Total: 4 Question Answer Marks Guidance 2(ii)(b) Max velocity at lowest point [0.2 × 10 × 6 = 0.5 × 0.2 × (v 2 – 5 2)] M1 PE loss = KE gain v 2 = 145 → v = 12(.0) ms–1 A1 Total: 2

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3 A particle A moves in a straight line with constant speed 10 m s−1. Two seconds after A passes a point O on the line, a particle B passes through O, moving along the line in the same direction as A. Particle B has speed 16 m s−1 at O and has a constant deceleration of 2 m s−2. (i) Find expressions, in terms of t, for the displacement from O of each particle t s after B passes through O. 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(ii) Find the distance between the particles when B comes to instantaneous rest. 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(iii) Find the minimum distance between the particles. 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Mark scheme: 3(i) M1 Attempt sA as sA = k + 10t (any k) sA = 20 + 10t A1 sB = 16t + ½(–2)t 2 [= 16t – t 2] B1 FT Allow FT only if sA = 10t and sB = 16(t – 2) + ½(–2)(t – 2) 2 i.e. t measured from when A passes O Total: 3 3(ii) vB = 16 – 2t → vB = 0, t = 8 B1 s = sA – sB [= 20 + 10t + t2 – 16t = t 2 – 6t + 20] M1 Finding distance between A and B at time t = T ( T > 0 ) found from a valid method for vB = 0 t = 8, s = 36 (m) A1 Total: 3 Question Answer Marks Guidance 3(iii) d 2 6 d = − s t t or s = t 2 – 6t + 20 = (t – 3) 2 + 11 M1 Either use differentiation or complete the square, or state value of t when speeds are the same [t = 3] M1 Solve for t and evaluate sA – sB at this value of t s = sA – sB = 11 m A1 Total: 3 [P = 850 × 42]

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4 A car of mass 1200 kg is moving on a straight road against a constant force of 850 N resisting the motion. (i) On a part of the road that is horizontal, the car moves with a constant speed of 42 m s−1. (a) Calculate, in kW, the power developed by the engine of the car. 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(b) Given that this power is suddenly increased by 6 kW, find the instantaneous acceleration of the car. 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(ii) On a part of the road that is inclined at 1Å to the horizontal, the car moves up the hill at a constant speed of 24 m s−1, with the engine working at 80 kW. Find 1. 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Mark scheme: 4(i)(a) M1 Using P = Fv P = 35700 W = 35.7 kW A1 Must be in kW to 3sf Total: 2 4(i)(b) P = 41700 → [DF = 41700/42] M1 Find new power and new DF based on power found in 4(i)(a) [(993 – 850) = 1200a] M1 Apply Newton 2, three terms a = 5/42 = 0.119 ms–2 A1 Total: 3 Question Answer Marks Guidance 4(ii) DF = 80000/24 B1 DF = P/v [DF – 850 – mg sin θ = 0] M1 Newton 2 along the hill, 3 terms [12000 sin θ = 80000/24 – 850] θ = ….. M1 Solve for θ, from a three term equation θ = 11.9 A1 Total: 4

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5 P N 30Å 40Å A particle of mass 0.12 kg is placed on a plane which is inclined at an angle of 40Å to the horizontal. The particle is kept in equilibrium by a force of magnitude P N acting up the plane at an angle of 30Å above a line of greatest slope, as shown in the diagram. The coefficient of friction between the particle and the plane is 0.32. Find the set of possible values of P. 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Mark scheme: 5 M1 Resolve perpendicular to the plane, three terms R + P sin 30 = 0.12g cos 40 A1 R does not need to be the subject F = 0.32R M1 Use F = µR [Pmin cos 30 + F = 0.12g sin 40] M1 About to slip down, 3 terms [Pmax cos 30 – F = 0.12g sin 40] M1 About to slip up, 3 terms [P cos 30 = 0.12g sin 40 ±0.32 (0.12g cos 40 – P sin 30)] OR [P cos 30 ± 0.32R = 0.12g sin 40 R + P sin 30 = 0.12g cos 40] Must reach P =… in either method M1 Substitute for F and solve for P in either case, 4 terms OR solve a pair of simultaneous equations (each with 3 terms) in R and P for P in one of the cases Pmax = 1.04 Pmin = 0.676 A1 For either correct 0.676 ⩽ P ⩽ 1.04 A1 Total: 8

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6 A P B 1Å The diagram shows a fixed block with a horizontal top surface and a surface which is inclined at an angle of 1Å to the horizontal, where sin 1 = 35. A particle A of mass 0.3 kg rests on the horizontal surface and is attached to one end of a light inextensible string. The string passes over a small smooth pulley P fixed at the edge of the block. The other end of the string is attached to a particle B of mass 1.5 kg which rests on the sloping surface of the block. The system is released from rest with the string taut. (i) Given that the block is smooth, find the acceleration of particle A and the tension in the string. 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(ii) It is given instead that the block is rough. The coefficient of friction between A and the block is - and the coefficient of friction between B and the block is also -. In the first 3 seconds of the motion, A does not reach P and B does not reach the bottom of the sloping surface. The speed of the particles after 3 s is 5 m s−1. Find the acceleration of particle A and the value of -. 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Mark scheme: 6(i) A [T = 0.3a] B [1.5g sin θ – T = 1.5a] System [1.5g sin θ = 1.8a] M1 Apply Newton’s second law to A or to B or to the system A1 Any two correct equations M1 Solve 2 simultaneous equations for a and/or T or use the system equation. a = 9/1.8 = 5 ms–2 A1 T = 1.5 N A1 Total: 5 Question Answer Marks Guidance 6(ii) [5 = 3a] M1 v = u + at used with t = 3, u = 0, v = 5 a = 5/3 = 1.67 A1 RA = 3 RB = 15 cos 36.9 = 12 B1 For either reaction [FA = 3µ FB = 12µ] M1 Use F = µR for either term EITHER: A [T – FA = 0.3a] B [15 sin 36.9 – T – FB = 1.5a] System equation is [1.5g sin 36.9 – FA – FB = 1.8a] (M1 Apply Newton’s second law to A or to B or to the system A2/1/0 A1 Correct equation for A or B A2 Correct equations for A and B OR A2 Correct system equation [9 – 15µ = 3] M1 Solve for µ from equations with correct number of terms µ = 0.4 = 2/5 A1) OR: s = ½ (5/3) × 32 = 7.5 (B1 Find distance travelled in 3 secs PE loss = 1.5 × 10 × 7.5 × (3/5) = 67.5 B1 KE gain = ½ (1.8) × 52 = 22.5 B1 [67.5 = 22.5 + 3µ × 7.5 + 12µ × 7.5] M1 Use Work/Energy equation µ = 2/5 = 0.4 A1) Total: 9

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